A) \[1.006V\]
B) \[0.503V\]
C) \[0.235V\]
D) \[-0.235V\]
Correct Answer: C
Solution :
Cell reaction: \[2F{{e}^{3+}}+2{{I}^{-}}\to 2F{{e}^{2+}}+{{I}_{2}}\]\[F{{e}^{3+}}\]shows reduction while I shows oxidation (reduction occurs at cathode while oxidation takes place at anode) \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] Given, \[E_{cathode}^{o}(F{{e}^{3+}}/F{{e}^{2+}})=0.771V\] and\[E_{anode}^{\text{o}}({{I}^{-}}/{{I}_{2}})=0.536V\] \[\therefore \]\[E_{cell}^{o}=0.771-0.536\] \[=0.235V\]You need to login to perform this action.
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